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Naily [24]
3 years ago
14

A graphic designer charges an hourly rate. In July, the only expense is for advertising. The function graphed models the designe

r's net income, y, for July for x hours of work.
Select True or False to describe each statement.

I guessed on this question, but I'm not 100% sure. Please help! 

Mathematics
2 answers:
Art [367]3 years ago
8 0

Actually you have the answers correct I took the test.

ICE Princess25 [194]3 years ago
5 0
<span>The first one is false, but the other two are true.
</span>
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HELP NEEDED HERE ASAP​
svetoff [14.1K]

Answer:

5

Step-by-step explanation:

Plug in 6 in place of x in the expression

1/2x+2

1/2(6)+2

3+2

5

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Using a Line to Find a Point-Slope Equation and the Equivalent Function A coordinate grid with a line passing through the points
RoseWind [281]

Answer:

y = x/4 -1/2

Step-by-step explanation:

given coordinates : ( -2, -1 ) and ( 2 , 0 )

gradient = y2 - y1 / x2 - x1

               = 0 - -1 / 2 - -2

               = 1/4

equation of line:

y - y1 = m( x - x1 )

y - 0 = 1/4 ( x - 2 )

y = x/4 -1/2

the line shown below to confirm:

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The function y = 3.50 x + 2 represents the total amount of money, y, saved over x weeks.
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B it is linear Bc it increases constantly. There is a constant slope that doesn’t change
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3 years ago
Data collected at Toronto Pearson International Airport suggests that an exponential distribution with mean value 2725hours is a
Ivan

Answer:

a) What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

b) What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

P(X

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

P(X=x)=\lambda e^{-\lambda x}

The cumulative distribution for this function is given by:

F(X) = 1- e^{-\lambda x}, x\ geq 0

We know the value for the mean on this case we have that :

mean = \frac{1}{\lambda}

\lambda = \frac{1}{Mean}= \frac{1}{2.725}=0.367

Solution to the problem

Part a

What is the probability that the duration of a particular rainfall event at this location is at least 2 hours?

We want this probability"

P(X >2) = 1-P(X\leq 2) = 1-(1- e^{-0.367 *2})=e^{-0.367 *2}= 0.48

At most 3 hours?

P(X \leq 3) = F(3) = 1-e^{-0.367*3}= 1-0.333 =0.667

Part b

What is the probability that rainfall duration exceeds the mean value by more than 2 standard deviations?

The variance for the esponential distribution is given by: Var(X) =\frac{1}{\lambda^2}

And the deviation would be:

Sd(X) = \frac{1}{\lambda}= \frac{1}{0.367}= 2.725

And the mean is given by Mean = 2.725

Two deviations correspond to 5.540, so we want this probability:

P(X > 2.725 + 2*5.540) = P(X>13.62) = 1-P(X

What is the probability that it is less than the mean value by more than one standard deviation?

For this case we want this probablity:

P(X

8 0
3 years ago
In the context of correlational research, if there is no relationship between two variables, what is the correlation coefficient
mel-nik [20]

Answer:

The Correlation coefficient is 0

Step-by-step explanation:

3 0
3 years ago
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